{
  "nbformat": 4,
  "nbformat_minor": 0,
  "metadata": {
    "kernelspec": {
      "display_name": "Python 3 (ipykernel)",
      "language": "python",
      "name": "python3"
    },
    "language_info": {
      "codemirror_mode": {
        "name": "ipython",
        "version": 3
      },
      "file_extension": ".py",
      "mimetype": "text/x-python",
      "name": "python",
      "nbconvert_exporter": "python",
      "pygments_lexer": "ipython3",
      "version": "3.8.12"
    },
    "colab": {
      "name": "разбор ДЗ 5.ipynb",
      "provenance": []
    }
  },
  "cells": [
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "XAv9cZVp9N5x"
      },
      "source": [
        "# Линейные преобразования"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "3GDQO24h9N6M"
      },
      "source": [
        "# Домашнее задание"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "rtL9fAAS9N6P"
      },
      "source": [
        "__1.__ Найти собственные векторы и собственные значения для линейного оператора, заданного матрицей\n",
        "\n",
        "### $$A=\\begin{pmatrix}\n",
        "-1 & -6\\\\ \n",
        "2 & 6\n",
        "\\end{pmatrix}.$$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "F-uYq4N-9N6T"
      },
      "source": [
        "__Решение__"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "f1P_Aslo9N6W"
      },
      "source": [
        "Найдем собственные значения:"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "uuSV_YLM9N6X"
      },
      "source": [
        "###  $$\\begin{vmatrix}\n",
        "-1-\\lambda & -6\\\\ \n",
        "2 & 6-\\lambda\n",
        "\\end{vmatrix} = 0,$$\n",
        "\n",
        "### $$(-1-\\lambda)(6-\\lambda)-2\\cdot(-6)=0,$$\n",
        "\n",
        "### $$\\lambda^{2}-5\\cdot\\lambda+6=0,$$\n",
        "\n",
        "### $$D=25-24=1,$$\n",
        "\n",
        "### $$\\lambda_{1}=2, \\lambda_{2}=3.$$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "7G66zOx29N6a"
      },
      "source": [
        "Найдем собственный ветор для $\\lambda_{1}$:"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "rSn5dT5F9N6c"
      },
      "source": [
        "### $$\\begin{pmatrix}\n",
        "-1 & -6\\\\ \n",
        "2 & 6\n",
        "\\end{pmatrix}\n",
        "\\begin{pmatrix}\n",
        "x_{1}\\\\ \n",
        "x_{2}\n",
        "\\end{pmatrix}=\n",
        "2\n",
        "\\begin{pmatrix}\n",
        "x_{1}\\\\ \n",
        "x_{2}\n",
        "\\end{pmatrix},\n",
        "$$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "3yjaDn6x9N6g"
      },
      "source": [
        "### $$\\begin{cases}\n",
        "-x_{1}-6\\cdot x_{2}=2\\cdot x_{1}, \\\\\n",
        "2\\cdot x_{1}+6\\cdot x_{2}=2\\cdot x_{2},\n",
        "\\end{cases}$$ "
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "vwcYuFyo9N6j"
      },
      "source": [
        "### $$\\begin{cases}\n",
        "-3\\cdot x_{1}= 6\\cdot x_{2}, \\\\\n",
        "2\\cdot x_{1} = -4\\cdot x_{2},\n",
        "\\end{cases}$$ "
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "_QMzGVYV9N6q"
      },
      "source": [
        "__Ответ__"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "7478MC519N6r"
      },
      "source": [
        "###  $$x_{1}= -2\\cdot x_{2},$$\n",
        "\n",
        "Если $x_{2}=1$, то $x_{1}=-2$. Получаем вектор $(-2,1)$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "BMTen1229N6t"
      },
      "source": [
        "Найдем собственный ветор для $\\lambda_{2}$:"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "WG8fkFWq9N6y"
      },
      "source": [
        "### $$\\begin{pmatrix}\n",
        "-1 & -6\\\\ \n",
        "2 & 6\n",
        "\\end{pmatrix}\n",
        "\\begin{pmatrix}\n",
        "x_{1}\\\\ \n",
        "x_{2}\n",
        "\\end{pmatrix}=\n",
        "3\n",
        "\\begin{pmatrix}\n",
        "x_{1}\\\\ \n",
        "x_{2}\n",
        "\\end{pmatrix},\n",
        "$$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "U8p_oaL99N6z"
      },
      "source": [
        "### $$\\begin{cases}\n",
        "-x_{1}-6\\cdot x_{2}=3\\cdot x_{1}, \\\\\n",
        "2\\cdot x_{1}+6\\cdot x_{2}=3\\cdot x_{2},\n",
        "\\end{cases}$$ "
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "B7AWAfQ69N60"
      },
      "source": [
        "### $$\\begin{cases}\n",
        "-4\\cdot x_{1}= 6\\cdot x_{2}, \\\\\n",
        "2\\cdot x_{1} = -3\\cdot x_{2},\n",
        "\\end{cases}$$ "
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "d-mFMLVx9N64"
      },
      "source": [
        "__Ответ__"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "vDl8CChr9N65"
      },
      "source": [
        "### $$2x_{1}= -3\\cdot x_{2},$$\n",
        "\n",
        "Если $x_{2}=2$, то $x_{1}=-3$. Получаем вектор $(-3,2)$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "8a4NbUJu9N67"
      },
      "source": [
        "__2.__ Дан оператор поворота на 180 градусов, задаваемый матрицей \n",
        "\n",
        "### $$A=\\begin{pmatrix}\n",
        "-1 & 0\\\\ \n",
        "0 & -1\n",
        "\\end{pmatrix}.$$\n",
        "\n",
        "Показать, что __любой__ вектор является для него собственным."
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "zbFVJBRS9N69"
      },
      "source": [
        "__Решение__"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "8-w7gly79N7C"
      },
      "source": [
        "Найдем собственные значения:\n",
        "\n",
        "### $$\\begin{vmatrix}\n",
        "-1-\\lambda & 0\\\\ \n",
        "0 & -1-\\lambda\n",
        "\\end{vmatrix} = 0,$$\n",
        "\n",
        "### $$(-1-\\lambda)(-1-\\lambda)=0,$$\n",
        "\n",
        "### $$(-1-\\lambda)^{2}=0,$$\n",
        "\n",
        "### $$\\lambda_{1}=\\lambda_{2}=-1$$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "d8wqWV_W9N7D"
      },
      "source": [
        "Предположим, что любой вектор $x$ является собственным вектором заданного линейного оператора, тогда должно существовать некоторое вещественное число $\\lambda$, при котором \n",
        "\n",
        "### $$\\begin{pmatrix}\n",
        "-1 & 0\\\\ \n",
        "0 & -1\n",
        "\\end{pmatrix}\n",
        "\\begin{pmatrix}\n",
        "x_{1}\\\\ \n",
        "x_{2} \n",
        "\\end{pmatrix}=\n",
        "\\lambda\n",
        "\\begin{pmatrix}\n",
        "x_{1}\\\\ \n",
        "x_{2}\n",
        "\\end{pmatrix},\n",
        "$$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "cBV6ddC_9N7E"
      },
      "source": [
        "Из этого будет следовать, что \n",
        "\n",
        "### $$\\begin{cases}\n",
        "-x_{1}=\\lambda \\cdot x_{1}\\\\ \n",
        "-x_{2}=\\lambda \\cdot x_{2}\n",
        "\\end{cases}\n",
        ".$$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "iA3cpfri9N7H"
      },
      "source": [
        "При $\\lambda = -1$ получаем:\n",
        "\n",
        "### $$\\begin{cases}\n",
        "-x_{1}=-x_{1}\\\\ \n",
        "-x_{2}=-x_{2}\n",
        "\\end{cases}$$\n",
        "\n",
        "или \n",
        "\n",
        "### $$\\begin{cases}\n",
        "0\\cdot x_{1}=0\\\\ \n",
        "0\\cdot x_{2}=0\n",
        "\\end{cases}$$\n",
        "\n",
        "То есть $x_1$ и $x_2$ любые"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "RQb7QBPB9N7I"
      },
      "source": [
        "__3.__ Пусть линейный оператор задан матрицей\n",
        "\n",
        "### $$A=\\begin{pmatrix}\n",
        "1 & 1\\\\ \n",
        "-1 & 3\n",
        "\\end{pmatrix}.$$\n",
        "\n",
        "Установить, является ли вектор $x=(1,1)$ собственным вектором этого линейного оператора."
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "U3y-vybA9N7I"
      },
      "source": [
        "__Решение__"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "to-0HRXk9N7J"
      },
      "source": [
        "Предположим, что вектор $x$ является собственным вектором заданного линейного оператора, тогда должно существовать некоторое вещественное число $\\lambda$, при котором \n",
        "\n",
        "### $$\\begin{pmatrix}\n",
        "1 & 1\\\\ \n",
        "-1 & 3\n",
        "\\end{pmatrix}\n",
        "\\begin{pmatrix}\n",
        "1\\\\ \n",
        "1 \n",
        "\\end{pmatrix}=\n",
        "\\lambda\n",
        "\\begin{pmatrix}\n",
        "1\\\\ \n",
        "1\n",
        "\\end{pmatrix}.\n",
        "$$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "YULTOs3U9N7J"
      },
      "source": [
        "Из этого будет следовать, что \n",
        "\n",
        "### $$\\begin{cases}\n",
        "1+1=1\\cdot\\lambda \\\\ \n",
        "-1+3 = 1\\cdot \\lambda\n",
        "\\end{cases}\n",
        "\\Rightarrow\n",
        "\\begin{cases}\n",
        "\\lambda=2\\\\ \n",
        "\\lambda=2\n",
        "\\end{cases}.$$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "wwuqbl6R9N7L"
      },
      "source": [
        "Таким образом, вектор $x=(1,1)$ является собственным вектором линейного оператора, заданного матрицей $A$, и его собственное значение составляет $\\lambda=2.$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "q6qOea1U9N7N"
      },
      "source": [
        "__4.__ Пусть линейный оператор задан матрицей\n",
        "\n",
        "### $$A=\\begin{pmatrix}\n",
        "0 & 3 & 0\\\\ \n",
        "3 & 0 & 0\\\\\n",
        "0 & 0 & 3\n",
        "\\end{pmatrix}.$$\n",
        "\n",
        "Установить, является ли вектор $x=(3, -3, -4)$ собственным вектором этого линейного оператора."
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "hT6Dg-jT9N7P"
      },
      "source": [
        "__Решение__"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "W4VO16z09N7P"
      },
      "source": [
        "Предположим, что вектор $x$ является собственным вектором заданного линейного оператора, тогда должно существовать некоторое вещественное число $\\lambda$, при котором \n",
        "\n",
        "### $$\\begin{pmatrix}\n",
        "0 & 3 & 0\\\\ \n",
        "3 & 0 & 0\\\\\n",
        "0 & 0 &3\n",
        "\\end{pmatrix}\n",
        "\\begin{pmatrix}\n",
        "3\\\\ \n",
        "-3\\\\\n",
        "-4\n",
        "\\end{pmatrix}=\n",
        "\\lambda\n",
        "\\begin{pmatrix}\n",
        "3\\\\ \n",
        "-3\\\\\n",
        "-4\n",
        "\\end{pmatrix}.\n",
        "$$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "zGYthXFI9N7Q"
      },
      "source": [
        "Из этого будет следовать, что \n",
        "\n",
        "### $$\\begin{cases}\n",
        "-9=3\\cdot\\lambda \\\\ \n",
        "9 = -3\\cdot \\lambda \\\\\n",
        "-12=-4\\cdot\\lambda\n",
        "\\end{cases}\n",
        "\\Rightarrow\n",
        "\\begin{cases}\n",
        "\\lambda=-3\\\\ \n",
        "\\lambda=-3 \\\\\n",
        "\\lambda=3\n",
        "\\end{cases}.$$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "y9rbJw2h9N7Q"
      },
      "source": [
        "Такая система не имеет смысла, следовательно, вектор $x=(3, -3, -4)$ не является собственным вектором линейного оператора, заданного матрицей $A$."
      ]
    }
  ]
}